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Talk:Fate of the Demonfolk/@comment-27867683-20170330103050/@comment-30176423-20170330180944
I see. Because the initial combines used 4 copies at a time, it is difficult to see when did the "CR-1 -> CR-2" and "CR-2 -> CR-3" upgrade happened. Still you got up to CR-3 reasonably fast. And then got stuck there with 17 consecutive failures. Assuming a classic "100% -> 50% -> 25% -> 25% -> 25%" cost reduction model, the last streak of consecutive failures had a 0.75 ^ 17 = ~0.75% chance of happening. And using the remaining 12 copies for combining, you now can reach CR-4 (97% chance) or CR-5 (84% chance). Of course, this is only valid if the classic cost reduction model is really applicable in your case. Assuming a modified model with "100% -> p -> p/2 -> p/2 -> p/2" cost reduction probabilities, we first need to estimate the value of 'p': * On one hand, you had a fast mincosting to CR-3 using 8 copies (9 copies including the base unit). And on average you would need to merge 9 copies to reach CR-3 when p=43%. On the other hand, the 17 consecutive failures in the end would have a (1 - 0.43/2) ^ 17 = ~1.6% chance. Using the remaining 12 copies for combining, you now can reach CR-4 (94.5% chance) or CR-5 (76.5% chance). But we kind of combine average luck in the beginning with very bad luck in the end for the 'p' value estimate, which is probably very much off. * On the other hand, you had one successful cost reduction in the last 21 attempts (4 + 17). This successful cost reduction could have happened on any of these attempts, so let's "balance" the luck in the beginning and in the end. So, just for the sake of the 'p' parameter estimation, we can pretend that CR-3 happened using 19 copies and then you had 7 failures in the end. Reaching CR-3 with 19 copies on average corresponds to p=17.6%. Then having 7 failures in a row has a ~52% chance, which is also average luck. In my opinion, this is probably the most reasonable estimate that we can get from your data. Now if you try to merge your remaining 12 copies, each attempt would have a p/2=8.8% chance of success and you can reach CR-4 (67% chance) or CR-5 (28.5% chance). I got these numbers using the online binomial calculator (Probability of success on a single trial: 0.088, Number of trials: 12, Number of successes: 1 for CR-4 or 2 for CR-5). To sum it up, if you continue trying to mincost your current copy with 12 more copies (and keep or improve the already high 7/10 skill level), then: * The classic model predicts CR-4 (97% chance) or CR-5 (84% chance) results. * My modified model with an estimated parameter p=17.6% from your data predicts CR-4 (67% chance) or CR-5 (28.5% chance) results. CR-4 chance there means CR-4 or better (so it includes CR-5 case too). You have a roughly 1 out of 3 chance to remain at CR-3 too.